Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
//package TIM;


import java.io.*;
import java.lang.*;
import java.math.*;
import java.util.*;


public class Main {
    static Scanner input = new Scanner(System.in);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(System.out);
    static long MOD = 1000000007;
    static ArrayList<Integer>[] adjacent;

    public static void main(String[] args) throws IOException {
        int n = input.nextInt();
        int m = input.nextInt();
        int[][] a = new int[n + 1][m + 1];
        int[][] b = new int[n + 1][m + 1];
        int num = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                a[i][j] = input.nextInt();
                b[i][j]=a[i][j];
            }
        ArrayList<String> ans = new ArrayList<String>();
        for (int i = 0; i < n; i++) {
            int min = a[i][0];
            for (int j = 1; j < m; j++)
                if (a[i][j] < min) min = a[i][j];
            if (min == 0) continue;
            for (int j = 0; j < min; j++)
                ans.add("row " + (i + 1));
            for (int j = 0; j < m; j++)
                a[i][j] -= min;
        }
        for (int j = 0; j < m; j++) {
            int min = a[0][j];
            for (int i = 1; i < n; i++)
                if (a[i][j] < min) min = a[i][j];
            if (min == 0) continue;
            for (int i = 0; i < min; i++)
                ans.add("col " + (j + 1));
            for (int i = 0; i < n; i++)
                a[i][j] -= min;
        }
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (a[i][j] != 0) {
                    System.out.println(-1);
                    return;
                }
        ArrayList<String> ans2 = new ArrayList<String>();
        for (int j = 0; j < m; j++) {
            int min = b[0][j];
            for (int i = 1; i < n; i++)
                if (b[i][j] < min) min = b[i][j];
            if (min == 0) continue;
            for (int i = 0; i < min; i++)
                ans2.add("col " + (j + 1));
            for (int i = 0; i < n; i++)
                b[i][j] -= min;
        }
        for (int i = 0; i < n; i++) {
            int min = b[i][0];
            for (int j = 1; j < m; j++)
                if (b[i][j] < min) min = b[i][j];
            if (min == 0) continue;
            for (int j = 0; j < min; j++)
                ans2.add("row " + (i + 1));
            for (int j = 0; j < m; j++)
                b[i][j] -= min;
        }
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (b[i][j] != 0) {
                    System.out.println(-1);
                    return;
                }
        if (ans.size() < ans2.size()) {
            System.out.println(ans.size());
            for (String s : ans)
                System.out.println(s);
            return;
        }
        System.out.println(ans2.size());
        for (String s : ans2)
            System.out.println(s);
    }
}